The frequency n of vibration of a stretched string is a function of its tension F, the length l and the mass per unit length m. From the knowledge of dimensions, prove that, n∝1/l √f/m Pls fast

n ∝ (1/l) (√F/m)

Step-by-step explanation:

Since frequency, n of vibration of a stretched string is a function of its tension F, the length l and the mass per unit length m, and has dimensions [T]⁻¹ where T = time, its dimension must be equal to that of the combination of F, L and m

Since n = f(F,L,m)

n = kFᵃLᵇmˣ

dimension of n = (dimension of F)ᵃ × (dimension of L)ᵇ × (dimension of m)ˣ

Since n = frequency, dimension of n = [T]⁻¹

F = Force, dimension of F = [M][L][T]⁻²

Also, L = length, dimension of L = [L] and

m = mass per unit length, dimension of m = [M][L]⁻¹

So, n = FᵃLᵇmˣ

[T]⁻¹  = ([M][L][T]⁻²)ᵃ( [L] )ᵇ([M][L]⁻¹)ˣ

[T]⁻¹  = ([M]ᵃ[L]ᵃ[T]⁻²ᵃ[L]ᵇ([M]ˣ[L]⁻ˣ

[M]⁰[L]⁰[T]⁻¹  = ([M]ᵃ ⁺ˣ)([L]ᵃ ⁺ ᵇ ⁻ˣ)([T]⁻²ᵃ)

equating the exponents on both sides, we have

a + x = 0  (1 )⇒ x = -a

a + b - x = 0 (2) ⇒

-2a = -1 (3) ⇒ a = 1/2

Substituting x into 2, we have

a + b -(-a) = 0

a + b + a = 0

2a + b = 0

b = -2a

b = -2 (1/2)

b = -1

x = -a = -1/2

So, substituting the variables into n, we have

n = kFᵃLᵇmˣ

n = kFᵃLᵇmˣ

n = k/l(√F/m)

n ∝ (1/l) (√F/m)

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph